# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Slide Rule Azimuth**

**From:**Gary LaPook

**Date:**2009 May 31, 00:29 -0700

The formula is very convenient for use with a slide rule in conjunction with the sine-cosine method and is the one I used to illustrate this computation. A while ago I posted the Rust diagram for finding azimuth which is in Weem's LOP Book. The Rust diagram was computed with this formula so it suffers from the same ambiguity. To deal with this Rust provides an auxiliary diagram that allows you to determine if the body is north or south of east. Here is a link to the Rust diagrams, http://fer3.com/arc/img/103383.rust%20diagram.pdf So, print out a copy and carry it with your calculator or slide rule. gl George Huxtable wrote: > Greg Rudzinski wrote, in [8443]- > > An interesting azimuth formula presented by H.H. Shufeldt in his book > SLIDE RULE FOR THE MARINER (pg. 77) > > Azimuth = INV SIN of COS declination Sin meridian angle divided by > COS altitude ( Ho or Hc ) > > Shufeldt states that Ho or Hc altitudes can be used. The INV SIN > result is added or subtracted from 360 or 180 degrees depending on > orientation. > > An alternate arrangement for the formula: > > Azimuth = INV SIN of SEC altitude COS declination SIN meridian angle > > I like the expediency of this formula but it does suffer from > inadequate slide rule scale resolution for azimuths approaching 270 > or 90 degrees. A trick to by-pass this problem for a sun observation > would be to directly observe a corrected bearing of the sun (which > should be low in the sky) for use as an altitude intercept azimuth. > > and Gary LaPook responded- > > That is the formula that I have used for years for calculating azimuth. > You can find it in Bowditch. George has pointed out that it gets > ambiguous near east and west but it is not a problem in real life and is > quick and easy to do on a calculator or slide rule. For those rare cases > near east or west another formula could be use. The Az calculated with > this formula is between zero and ninety degrees so you have to figure > what quadrant you are in and convert to Zn but this is also not a > problem in real life since you know the approximate direction when you > pointed your sextant. See: > | > | > http://groups.google.com/group/NavList/browse_thread/thread/af4f15cde5075f8f/058fe8755eeaca37?hl=en&lnk=gst&q=lapook+cosine#058fe8755eeaca37 > | > | > http://groups.google.com/group/NavList/browse_thread/thread/529edc05997d59d7/e002865149e31596?hl=en&lnk=gst&q=lapook+cosine#e002865149e31596 > > ================================ > > This question has been around this list, and its predecessor, more than > once, but it might as well get another airing. > > Gary has pointed out the ambiguity, for azimuths near East and West, which > is the serious drawback to this method of working (more serious, in its way, > that the poor precision at these angles, which Greg did recognise). But he > pointed it out, only to dismiss it, as "not a problem in real life". I > suggest he should think again. The fact that it may be "quick and easy to do > on a calculator or slide rule" does not overcome those difficulties > > He refers to those "rare cases" when the object is near East or West. Not so > rare, however. In the tropics, there are two periods of the year when the > Sun is either nearly-East or nearly-West, the whole day through. Elsewhere, > it's always near East-West twice a day, in Summer, just the best time for > determining longitude. > > The difficulty is that it's impossible to distinguish, by this method, > between azimuths greater than 90�, and azimuths correspondingly less than > 90�, such as between azimuths of 80� and 100�, as their sines are exactly > the same. As long as those azimuths differ sufficiently from 90�, there's no > problem; it's obvious which is the right value. Perhaps Gary is confident of > his ability to distinguish between azimuths of 80� and 100�, but could he do > so, in rough weather, for a high sky-object that might be 85�, or might be > 95�? If he got that choice wrong, the resulting 10� of error could upset a > position calculation, unless the intercept happened to be a short one. > > Gary suggests that in such cases, a navigator could use a different formula, > as indeed he could. But that means he would have to keep two different > procedures in his mental locker, and know when to apply each one. How much > simpler, then, to use instead a formula that always preserves its accuracy > over all azimuths, and is free from ambiguity. This is the formula that > derives azimuth from its tan, rather than sin or cos, as follows- > > Azimuth from North = arc tan ( sin (MA) / (cos lat tan dec -cos (MA) sin > lat)) > > If the result is negative, add 180 degrees to make it positive. This is how > it works if, like many navigators, you always think of your meridian angle > as a positive quantity, whether it's East ot West. That result would be the > azimuth of a body if it's East of you. If the body is to your West, the > angle from North would be the same, but measured from North the other way, > in the Western hemisphere, so you have to subtract that result from 360�. > > Personally, I prefer to think of meridian angles (and longitudes) as > increasing Westerly, just as Hour angles do (and against the current > conventions), in whch case the rules for getting the angle in the right > quadrant are a bit different. > > Although this method may take a few more keystrokes on a calculator, it has > the advantage that it doesn't depend on the result of any previous > calculation, for altitude. > > George. > > contact George Huxtable, at george@hux.me.uk > or at +44 1865 820222 (from UK, 01865 820222) > or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. > > > > > > > > > > > > --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---